Réponse :
Explications :
To solve this problem, we can use the formulas for calculating current, power, and apparent power (in kVA) in a three-phase AC system.
Given:
Voltage (V) = 400 V
Resistance (R) = 30 ohms (for each phase)
Inductive reactance (X) = 40 ohms (for each phase)
a) Current supplied by the alternator:
The current in each phase can be calculated using the formula:
�
=
�
�
I=
Z
V
Where:
�
V is the voltage (400 V in this case)
�
Z is the impedance, given by
�
2
+
�
2
R
2
+X
2
For this problem, we have a delta-connected load, so the impedance is the sum of resistance and reactance:
�
=
�
+
�
�
Z=R+jX
�
=
30
+
�
40
Z=30+j40
Now, let's calculate the current:
�
=
3
0
2
+
4
0
2
=
900
+
1600
=
2500
=
50
Ω
Z=
30
2
+40
2
=
900+1600
=
2500
=50Ω
�
=
400
50
=
8
A
I=
50
400
=8A
b) Output power and apparent power (kVA) of the alternator:
The output power (P) in each phase can be calculated using the formula:
�
=
�
�
cos
(
�
)
P=VIcos(ϕ)
Where:
�
V is the voltage (400 V)
�
I is the current (8 A)
cos
(
�
)
cos(ϕ) is the power factor, which for a purely resistive load is 1.
The total output power in three phases is
3
×
�
3×P.
�
=
400
×
8
×
1
=
3200
W
P=400×8×1=3200W
Total output power
=
3
×
3200
=
9600
W
Total output power=3×3200=9600W
To calculate the apparent power (S) in kVA, we use the formula:
�
=
�
2
+
�
2
S=
P
2
+Q
2
Where:
�
P is the active power (in watts)
�
Q is the reactive power (in VAR)
For a purely resistive load,
�
=
0
Q=0.
�
=
320
0
2
+
0
2
=
3200
VA
=
3.2
kVA
S=
3200
2
+0
2
=3200VA=3.2kVA
So, the output power of the alternator is 9600 W and the apparent power is 3.2 kVA.
